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Unformatted text preview: Chapter 5 The First Law of Thermodynamics 5221 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and the two tanks come to the same state at the temperature of the surroundings. The final pressure and the amount of heat transfer are to be determined. Assumptions 1 The tanks are stationary and thus the kinetic and potential energy changes are zero. 2 The tank is insulated and thus heat transfer is negligible. 3 There are no work interactions. Analysis We take the entire contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as ] ) ( ) ( [ ] [ 0) = PE = KE (since ) ( ) ( 1 1 1 1 2 total 2, , 1 , 1 , 2 out out energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net out in B A B A B A B A u m u m u m U U U Q W U U U Q E E E = = = ∆ + ∆ = ∆ = ∆ = + The properties of water in each tank are (Tables A4 through A6) Tank A: kJ/kg 1949.3 , 31 . 604 /kg m 0.4625 , 001084 . 80 . kPa 400 3 1 1 = = = = = = fg f g f u u v v x P ( 29 [ ] ( 29 kJ/kg 2163.75 3 . 1949 8 . 31 . 604 /kg m 0.3702 001084 . 4625 . 8 . 001084 . 1 , 1 3 1 , 1 = × + = + = = × + = + = fg f A fg f A u x u u v x v v Tank B: kJ/kg 2304.9 , 88 . 104 /kg m 43.36 , 001003 . /kg m 0.731 C 25 /kg m 0.731 kg 0.957 m 0.7 kg 0.957 417 . 540 . kg 0.417 /kg m 1.1988 m 0.5 kg 0.540 /kg m 0.3702 m 0.2 kJ/kg 2731.2 /kg m 1.1988 C 250 kPa 200 3 3 2 2 3 3 2 , 1 , 1 3 3 , 1 , 1 3 3 , 1 , 1 , 1 3 , 1 1 1 = = = = = = = = = = + = + = = = = = = = = = = = fg f g f t t B A t B B B A A A B B u u v v v T m V v m m m v V m v V m u v T P Thus at the final state the system will be a saturated liquidvapor mixture since v f < v 2 < v g . Then the final pressure must be P 2 = P sat @ 25 ° C = 3.169 kPa Also, ( 29 kJ/kg 143.60 2304.9 0.0168 104.88 0168 . 001 . 36 . 43 001 . 731 . 2 2 2 2 = × + = + = = = = fg f fg f u x u u v v v x Substituting, Q out = [(0.957)(143.6)  (0.540)(2163.75)  (0.417)(2731.2)] = 2170 kJ 5208 H 2 O 400 kPa H 2 O 200 kPa × B A Q Chapter 5 The First Law of Thermodynamics 5222 Problem 5221 is reconsidered. The effect of the environment temperature on the final pressure and the heat transfer as the environment temperature varies from 0°C to 50°C is to be investigated. The final results are to be plotted against the environment temperature. "Knowns" Vol_A=0.2 "[m^3]" P_A[1]=400 "[kPa]" x_A[1]=0.8 T_B[1]=250 "[C]" P_B[1]=200 "[kPa]" Vol_B=0.5 "[m^3]" T_final=25 "[C]" "T_final = T_surroundings. To do the parametric study or to solve the problem when Q_out = 0, place this statement in {}." {Q_out=0"[kJ]"} "To determine the surroundings temperature that makes Q_out = 0, remove the {} and resolve the problem." "Solution" "Conservation of Energy for the combined tanks:"...
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This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.
 Spring '08
 Chung

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